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EXPLANATION OF LANTHANUM IONIZATIONS
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) June 11 , 2015 Lanthanum is a chemical element with symbol La and atomic number 57. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this correct image of Lanthanum including the following ground state electron configuration: 1s22s22p63s23p63d104s24p64d105s25px25py25pz2 5d16s2 According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of lanthanum (from (E1 to E5 ) are the following: E1 = 5.58 , E2 = 11, E3 = 19.18, E4 = 49.95, and E5 = 61.6 . For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper of 2008. EXPLANATION OF E1 = 5.58 eV = -E(6s2) + E(6s1) Here the E(6s2) represents the binding energy of 6s2, while the E(6s1) represents the binding energy of 6s1. The charges (-54e) of (1s22s22p63s23p63d104s24p64d105s25p6) screen the nuclear charge (+57e) and for a perfect screening we would have ζ = 3 Note that the 6s2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(6s2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 6s1 consists of one electron, we apply the Bohr formula to write E(6s1) = (-13.6057)ζ2/n2 Therefore E1 = 5.58 eV = -E(6s2) + E(6s1) = - (16.95)ζ + 4.1) / n2 Then using n = 6 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 196.78 = 0 and solving for ζ we get ζ = 4.48 > 3. This means that the two electrons of 6s2 along with the 5d1 penetrate the 5p6 leading to the deformation of electron clouds. Of course the two electrons of opposite spin (6s2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy, which seems to be like a simple electric repulsion of the Coulomb law. This situation of a vibration energy due to an electromagnetic interaction indeed occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT). However under the influence of invalid relativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that it is due to the Coulomb repulsion. Under such fallacious ideas I published my paper of 2008 EXPLANATION OF E2 = 11 eV = -E(6s1) As in the case of E1 the charges (-54e) of (1s22s22p63s23p63d104s24p64d105s25px25py25pz2) screen the nuclear charge (+57e) and for a perfect screening we would have an effective ζ = 3. However the one electron of 6s1 palong with the electron of the 5d1 apenetrate the 5p6 and lead to the deformation of electron clouds. Thus ζ > 3. Here the E(6s1) represents the binding energy of (6s1) given by applying the Bohr formula as E2 = 11 eV = -Ε(6s1) = - ( -13.6057)ζ2 /62 Then solving for ζ we get ζ = 5.4 > 3 . Here the ζ = 5.4 > 4.48 > 3 means that the one electron breaks the symmetry and leads to a greater deformation of electron clouds. EXPLANATION OF E3 = 19.18 eV = -E(5d1) Here the E(5d1) represents the binding energy of the one electron (5d1) given by applying the Bohr formula as E3 = 19.18 = -E(5d1) = -(-13.6057)ζ2 /n2 As in the cases of 6s the charges (-54e) of (1s22s22p63s23p63d104s24p64d105s25px25py25pz2) screen the nuclear charge (+57e) and for a perfect screening we would have an effective ζ = 3. However the one electron of 5d1 repels the electrons of 5p6 and leads to the deformation of electron clouds. Thus ζ > 3. Then using n = 5 we get ζ = 5.9 > 3. ' ' EXPLANATION OF E4 = 49.95 = '-E(5px2) + E(5px1)' Here the E(5px2) represents the binding energy of 5px2, while the E(5px1) represents the binding energy of 5px1. The charges (-48e) of (1s22s22p63s23p63d104s24p64d105s2) screen the nuclear charge (+57e) and for a perfect screening we would have ζ = 9. Note that the 5px2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(5px2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 5px1 consists of one electron, we apply the Bohr formula to write E(5px1) = (-13.6057)ζ2/n2 Therefore E4 = 49.95 eV = -E(5px2) + E(5px1) = - (16.95)ζ + 4.1) / n2 Then using n = 5 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 1244.65 = 0 and solving for ζ we get ζ = 10.2 > 9. Here ζ = 10.2 > 9 means that the electrons of 5p repel the electrons of 5s2 from symmetrical positions in order to provide an effective ζ = 10.2 > 9 which is a little greater than the perfect screening with ζ = 9 . EXPLANATION OF E5 = 61.6 eV = '-E(5py2) + E(5py1)' Here the E(5py2) represents the binding energy of 5py2, while the E(5py1) represents the binding energy of 5py1. As in the case of E4 the charges (-48e) of (1s22s22p63s23p63d104s24p64d105s2) screen the nuclear charge (+57e) and for a perfect screening we would have ζ = 9. Note that the 5py2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(5py2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 5py1 consists of one electron, we apply the Bohr formula to write E(5py1) = (-13.6057)ζ2 / n2 Therefore E5 = 61.6 eV = -E(5py2) + E(5py1) = - (16.95)ζ + 4.1) / n2 Then using n = 5 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 1535.9 = 0 and solving for ζ we get ζ = 11.27 > 9. Here ζ = 11.27 > 10.2 > 9 means that the electrons of 5p after the ionizations break the symmetry for providing an effective ζ = 11.27 > 10.2 . Category:Fundamental physics concepts